Question

b. given the initial population size of population a and assuming that the population is experiencing growth at the growth rate r (calculated above), what will the number of plants be in each of the next 5 years? (use the initial population size as time 0.) round each to the nearest whole number and record your answers here (no grids provided).

time (year)	population size
1
2
3
4
5

(adapted from: mastering biology, pearson 2013)

1. a population of 265 swans was introduced to circle lake. the populations birth rate is 0.341 swans/year per capita, and the death rate is 0.296 swans/year per capita. what is the rate of population growth per capita, and is it increasing or decreasing? round to the nearest thousandth.
2. there are 190 grey tree frogs in a swamp. the population is under carrying capacity. if r = (-0.093) frogs/year, predict the population size next year. round to the nearest whole number.
3. a population of 1492 baltimore orioles was introduced to an area of nerstrand woods. over the next year, the orioles show a death rate of 0.395 while the population drops to 1134. whats the birth rate for this population? round to the nearest hundredth.
4. there are 780 turkeys living in merriam township, which is 92 acres in size. the birth rate is 0.472 turkeys/year per capita. the death rate is 0.331 turkeys/year per capita.

a. what is the population density? round to the nearest tenth.
b. what is dn/dt? round to the nearest whole number.
c. predict n after one year, assuming dn/dt stays constant. round to the nearest whole number.

1. one dandelion plant can produce many seeds leading to a high growth rate for dandelion populations. if a population of dandelions is currently 40 individuals and r_max = 0.2 dandelions/month per capita, predict how many dandelions would be in this population after 4 months. round to the nearest whole number.
2. imagine the dandelion population of 40 (in #18) cannot continue to grow exponentially due to lack of space. the carrying capacity for their patch of lawn is 70 dandelions. what is their dn/dt in this logistic growth situation? round to the nearest tenth.
3. a hypothetical population has a carrying capacity of 1,500 individuals and r_max is 1.0. fill out the following table. round all answers to the nearest whole number and record your answers here (no grids provided).

population size	show work here	population growth
1750
2000

Explanation:

Step1: Recall population - growth formula

The exponential - growth formula is $N_t=N_0e^{rt}$, where $N_t$ is the population size at time $t$, $N_0$ is the initial population size, $r$ is the growth rate, and $t$ is the time in years. But if the growth rate is small, we can also use the simple formula $N_t = N_0(1 + r)^t$. Since the problem doesn't give the growth rate $r$ for the first part, we'll assume a general approach for population - growth problems.

Problem 14:

1. Calculate the per - capita growth rate:

• The per - capita growth rate $r$ is given by $r = b - d$, where $b$ is the birth rate and $d$ is the death rate.
• Given $b = 0.341$ swans/year per capita and $d=0.296$ swans/year per capita.
• $r=b - d=0.341−0.296 = 0.045$ swans/year per capita. Since $r>0$, the population is increasing.

Problem 15:

1. Predict the population size next year:

• The formula for population change is $N_{t + 1}=N_t+rN_t=N_t(1 + r)$.
• Given $N_t = 190$ and $r=-0.093$.
• $N_{t + 1}=190\times(1 - 0.093)=190\times0.907 = 172.33\approx172$ grey tree frogs.

Problem 16:

1. Find the birth rate:

• We know that $N_1=N_0+(b - d)N_0$. Rearranging for $b$, we get $b=\frac{N_1 - N_0}{N_0}+d$.
• Given $N_0 = 1492$, $N_1 = 1134$, and $d = 0.395$.
• First, calculate $\frac{N_1 - N_0}{N_0}=\frac{1134 - 1492}{1492}=\frac{- 358}{1492}\approx - 0.2399$.
• Then $b=-0.2399 + 0.395=0.1551\approx0.16$.

Problem 17:

1. Calculate population density:

• Population density $D=\frac{N}{A}$, where $N$ is the population size and $A$ is the area.
• Given $N = 780$ turkeys and $A = 92$ acres.
• $D=\frac{780}{92}\approx8.5$ turkeys/acre.

1. Calculate $\frac{dN}{dt}$:

• $\frac{dN}{dt}=(b - d)N$. Given $b = 0.472$ turkeys/year per capita, $d = 0.331$ turkeys/year per capita, and $N = 780$.
• First, $b - d=0.472-0.331 = 0.141$.
• Then $\frac{dN}{dt}=0.141\times780 = 109.98\approx110$.

1. Predict $N$ after one year:

• If $\frac{dN}{dt}$ is constant, then $N_{t+1}=N_t+\frac{dN}{dt}$.
• Given $N_t = 780$ and $\frac{dN}{dt}=110$.
• $N_{t + 1}=780 + 110=890$ turkeys.

Problem 18:

1. Predict the population size after 4 months:

• Using the formula $N_t=N_0(1 + r_{max}t)$. Given $N_0 = 40$, $r_{max}=0.2$ dandelions/month per capita, and $t = 4$ months.
• $N_t=40\times(1+0.2\times4)=40\times(1 + 0.8)=40\times1.8 = 72$ dandelions.

Problem 19:

1. Calculate $\frac{dN}{dt}$ in logistic growth:

• The formula for $\frac{dN}{dt}$ in logistic growth is $\frac{dN}{dt}=r_{max}N(1-\frac{N}{K})$.
• Given $N = 40$, $K = 70$, and assuming $r_{max}=0.2$ (from problem 18).
• $\frac{dN}{dt}=0.2\times40\times(1-\frac{40}{70})=8\times\frac{30}{70}=\frac{240}{70}\approx3.4$ dandelions/month.

Problem 20:

1. Calculate population growth in logistic growth:

• The formula for $\frac{dN}{dt}$ in logistic growth is $\frac{dN}{dt}=r_{max}N(1-\frac{N}{K})$. Given $r_{max}=1.0$ and $K = 1500$.
• For $N = 1600$:
• $\frac{dN}{dt}=1\times1600\times(1-\frac{1600}{1500})=1600\times(-\frac{100}{1500})\approx - 107$.
• For $N = 1750$:
• $\frac{dN}{dt}=1\times1750\times(1-\frac{1750}{1500})=1750\times(-\frac{250}{1500})\approx - 292$.
• For $N = 2000$:
• $\frac{dN}{dt}=1\times2000\times(1-\frac{2000}{1500})=2000\times(-\frac{500}{1500})\approx - 667$.

Answer:

1. $r = 0.045$, population is increasing
2. 172
3. 0.16
4. a. 8.5 turkeys/acre; b. 110; c. 890
5. 72
6. 3.4
7. For $N = 1600$, $\frac{dN}{dt}\approx - 107$; for $N = 1750$, $\frac{dN}{dt}\approx - 292$; for $N = 2000$, $\frac{dN}{dt}\approx - 667$